Question

(b): Finding probabilities for normal distribution

i. Given a population with a normal distribution, a mean of 0, and a standard deviation of 1, find the probability of a value less than 0.75.

ii. Given a population with a normal distribution, a mean of 120, and a standard deviation of 10, find the probability of a value greater than 135.

iii. Given a population with a normal distribution, a mean of 98.6, and a standard deviation of 0.62, find the probability of a value between 97.0 and 99.0.

iv. Given a population with a normal distribution, a mean of u = 17 and the standard deviation of 3, find the 85th percentile.

Answer 1

(b)

i.

$$P(Z<0.75)\approx0.773373$$

ii.

$$P(X>135)=1-P(X\leq 135)$$

$$=1-P(Z\leq\dfrac{135-120}{10})=1-P(Z\leq1.5)$$

$$\approx0.066807$$

iii.

$$P(97.0<X<99.0)=P(X<99.0)-P(X\leq 97.0)$$

$$=P(Z<\dfrac{99.0-98.6}{0.62})-P(Z\leq\dfrac{97.0-98.6}{0.62})$$

$$\approx P(Z<0.6451613)-P(Z\leq-2.5806452)$$

$$\approx0.7405889-0.0049308$$

$$\approx0.735658$$

iv.

$$P(Z<z)=0.85$$

$$z\approx1.036433$$

$$\dfrac{x-u}{\sigma}=1.036433$$

$$x=17+3(1.036433)$$

$$z\approx20.1093$$

The 85th percentile is $20.1093.$